For two projectiles with same speed to have the same horizontal range, angle of projection will be θ1+θ2=90∘.
Therefore, θ2=90∘−60∘=30∘.
Now, required sum of maximum heights will be,
(Hmax)1+(Hmax)2=2gu2sin2θ1+2gu2sin2θ2
=20402(41+43)=80m